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23-Chem-A3 Heat and Mass Transfer · May 2014

Question 6 of 7: Psychrometrics and Cooling-Tower Design

Nivaar worked solution (AI-drafted; not reviewed by a licensed engineer)

Notes on this paper

National Exams — May 2014 — 04-Chem-A3 Mass Transfer Operations. Three-hour, open-book exam; any non-communicating calculator permitted. Format: seven questions in three parts — answer one of Q1–Q2 (Part A), one of Q3–Q4 (Part B) and two of Q5–Q7 (Part C); four questions of equal value constitute a complete paper. All seven are solved below for completeness. Property data (diffusivities, vapour pressures, psychrometric enthalpies) are stated in each Given block; the psychrometric appendix supplied with the exam furnishes the saturated-air data used in Q6.

Reference texts: Geankoplis, Transport Processes and Separation Process Principles (4th ed., Prentice Hall) — molecular diffusion, convective mass transfer, absorption and humidification; Treybal, Mass-Transfer Operations (3rd ed., McGraw-Hill) — gas absorption, wetted-wall and cooling-tower design; Welty, Wicks, Wilson & Rorrer, Fundamentals of Momentum, Heat and Mass Transfer — boundary-layer and falling-sphere correlations; supporting property/psychrometric data from Perry's Chemical Engineers' Handbook (9th ed.).

Question 6: Psychrometrics and Cooling-Tower Design (equal value)

Question text not reproduced: the examination questions are © Engineers and Geoscientists BC. Open the official past paper (linked at the top of this page) to read the question, then follow the worked solution below.

Part (a) — Cool-then-reheat psychrometrics

Given. Moist air is cooled below its dew point (dropping liquid water) and then sensibly reheated at constant humidity. Saturation pressures come from the Antoine equation for water and match the supplied psychrometric chart.

QuantityValue
Initial state25 °C, 70% RH
Cooled to15 °C (below dew point ⇒ saturated)
Reheated to20 °C (constant humidity)
$p_{sat}$: 25/15/20 °C3.16 / 1.70 / 2.33 kPa

Find. (i) water removed, (ii)–(iii) wet-bulb temperatures, (iv) final RH, (v) final vapour pressure.

dry-bulb temperature T (°C)humidity 𝓗 (kg/kg)saturation (100% RH)1: 25 °C, 70% RH2: 15 °C, saturated3: 20 °C, reheated
Figure 6 — Air is cooled from state 1 (25 °C, 70% RH) down the saturation line to state 2 (15 °C, saturated), shedding moisture, then reheated at constant humidity to state 3 (20 °C).

Approach. Get the initial humidity from the RH and $p_{sat}(25)$; the air leaves the cooler saturated at 15 °C, so the humidity difference is the condensate; reheating is constant-humidity, fixing the final RH and vapour pressure; wet-bulb temperatures follow from the adiabatic-saturation relation.

  1. Initial humidity. $p_{w1} = 0.70\,p_{sat}(25) = 0.70(3.16) = 2.21$ kPa, so $\mathcal H_1 = \dfrac{0.622\,p_{w1}}{P-p_{w1}} = 0.01387$ kg/kg.
  2. (i) Water removed. Cooling to 15 °C saturates the air: $\mathcal H_2 = \mathcal H_{sat}(15) = 0.01060$ kg/kg. Condensate $= \mathcal H_1-\mathcal H_2 = \boxed{0.00327\ \text{kg/kg}\ (3.3\ \text{g/kg})}.$
  3. (iv)–(v) Reheat at constant humidity to 20 °C. $\mathcal H_3=\mathcal H_2$, so $p_{w3} = \dfrac{\mathcal H_3 P}{0.622+\mathcal H_3} = \boxed{1.70\ \text{kPa}}$ and $RH_3 = p_{w3}/p_{sat}(20) = 1.70/2.33 = \boxed{72.9\%}.$
  4. (ii) Initial wet-bulb. Solving the adiabatic-saturation relation $\mathcal H = \mathcal H_{sat}(T_w) - \dfrac{c_s}{\lambda}(T-T_w)$ at (25 °C, $\mathcal H_1$) gives $T_{w,1} = \boxed{21.0\ ^\circ\text{C}}.$
  5. (iii) Final wet-bulb. The same relation at (20 °C, $\mathcal H_3$) gives $T_{w,3} = \boxed{16.8\ ^\circ\text{C}}.$
QuantityResult
(i) Water removed3.3 g/kg dry air
(ii) Initial wet-bulb21.0 °C
(iii) Final wet-bulb16.8 °C
(iv) Final RH72.9%
(v) Final vapour pressure1.70 kPa

Part (b) — Cooling-tower packing height

Given. A counter-current cooling tower analysed by the Merkel enthalpy method; the supplied $k_ca$ (per second) is put on an enthalpy basis with the Lewis-number-one assumption.

QuantityValue
Water cooled35 → 20 °C, $L^{\prime}=0.556$ kg/m²·s
Air$G^{\prime}=0.833$ kg/m²·s, DB 20 °C, WB 15 °C
$k_ca$$2.95(L^{\prime})^{0.26}(G^{\prime})^{0.72}=2.22\ \text{s}^{-1}$
$(C_p)_{liq}$4.18 kJ/kg·K

Find. The packing height $Z$.

water temperature T (°C)enthalpy H (kJ/kg dry air)saturation H*(T)air in H₁=42air out H₂=8420253035
Figure 7 — Cooling-tower operating line (blue) from the air-inlet state at 20 °C to the outlet at 35 °C, kept below the saturated-air enthalpy curve $H^*(T)$ (green); the vertical gap is the enthalpy driving force.

Approach. Fix the inlet-air enthalpy from its wet-bulb temperature, set the operating line from the water/air energy balance, integrate the enthalpy transfer units, and multiply by the height of a transfer unit $H_{tOG}=G^{\prime}/(\rho_{air}k_ca)$.

  1. Air-inlet enthalpy. For air–water the wet-bulb is the adiabatic-saturation line, so $H_{G1}=H^*(15\,{}^\circ\text{C})\approx41.9$ kJ/kg.
  2. Operating line. Slope $L^{\prime}c_L/G^{\prime} = (0.556)(4.18)/0.833 = 2.79$ kJ/kg·°C, giving $H_{G2}=41.9+2.79(35-20)=83.7$ kJ/kg at the hot end.
  3. Enthalpy transfer units. $$N_{tOG}=\int_{20}^{35}\frac{(L^{\prime}c_L/G^{\prime})\,dT_L}{H^*(T_L)-H_G(T_L)} = 1.76,$$ evaluated by stepping $T_L$ and reading $H^*(T_L)$ from the saturated-air relation.
  4. Height of a transfer unit (Le = 1). Taking $k_ya\approx\rho_{air}k_ca = (1.185)(2.22)=2.63\ \text{s}^{-1}$-basis, $H_{tOG}=\dfrac{G^{\prime}}{\rho_{air}k_ca}=\dfrac{0.833}{2.63}=0.317$ m.
  5. Packing height. $$Z = H_{tOG}\,N_{tOG} = 0.317\times1.76 = \boxed{0.56\ \text{m}}.$$
QuantityResult
Inlet-air enthalpy $H_{G1}$41.9 kJ/kg
$N_{tOG}$1.76
$H_{tOG}$0.317 m
Packing height $Z$≈ 0.56 m
Check
The gas-phase coefficient supplied as $k_ca$ [s⁻¹] is converted to an enthalpy (mass) basis by assuming a Lewis number of one ($k_ya\approx\rho_{air}k_ca$), the standard air–water approximation. Saturated-air enthalpies use $H^*=(1.005+1.88\mathcal H)T+2501\mathcal H$, which reproduces the supplied psychrometric chart; only enthalpy differences enter $N_{tOG}$, so the datum cancels.